From d22cc2f282456f04e635a681e96efc54fb4ae405 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Thu, 25 Jan 2024 20:18:46 +0100 Subject: [PATCH] 14 --- inputs/lecture_13.tex | 4 +- inputs/lecture_14.tex | 185 ++++++++++++++++++++++-------------------- inputs/lecture_15.tex | 3 + 3 files changed, 103 insertions(+), 89 deletions(-) diff --git a/inputs/lecture_13.tex b/inputs/lecture_13.tex index c534efa..2a1cdb7 100644 --- a/inputs/lecture_13.tex +++ b/inputs/lecture_13.tex @@ -1,4 +1,5 @@ \lecture{13}{2023-11-08}{} + \gist{% % Recap $\LO = \{x \in 2^{\N\times \N} : x \text{ is a linear order}\} $. @@ -124,7 +125,8 @@ with $(f^{-1}(\{1\}), <)$. and $f$ is continuous. \end{proof} -% TODO: new section? +\subsection{$\Pi_1^1$-ranks} + \gist{% Recall that a \vocab{rank} on a set $C$ is a map $\phi\colon C \to \Ord$. diff --git a/inputs/lecture_14.tex b/inputs/lecture_14.tex index 3a52e99..b040edf 100644 --- a/inputs/lecture_14.tex +++ b/inputs/lecture_14.tex @@ -1,47 +1,52 @@ \lecture{14}{2023-12-01}{} -% TODO ANKI-MARKER \begin{theorem}[Moschovakis] If $C$ is coanalytic, then there exists a $\Pi^1_1$-rank on $C$. \end{theorem} \begin{proof} + \gist{% Pick a $\Pi^1_1$-complete set. It suffices to show that there is a rank on it. Then use the reduction to transfer it to any coanalytic set, - i.e.~for $x,y \in C'$ + i.e.~for $x,y \in C'$ let \[ x \le^{\ast}_{C'} y :\iff f(x) \le^\ast_{C} f(y) - \] + \] and similarly for $<^\ast$. - % https://q.uiver.app/#q=WzAsNSxbMCwwLCJZIl0sWzIsMCwiWCJdLFswLDEsIkMnIl0sWzIsMSwiQyJdLFsyLDIsIlxcUGlfMV4xLVxcdGV4dHtjb21wbGV0ZX0iXSxbMCwxLCJmIl0sWzIsMCwiXFxzdWJzZXRlcSIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzMsMSwiXFxzdWJzZXRlcSIsMix7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV1d -\[\begin{tikzcd} - Y && X \\ - {C'} && C \\ - && {\Pi_1^1-\text{complete}} - \arrow["f", from=1-1, to=1-3] - \arrow["\subseteq", hook, from=2-1, to=1-1] - \arrow["\subseteq"', hook, from=2-3, to=1-3] -\end{tikzcd}\] - + % https://q.uiver.app/#q=WzAsNCxbMCwwLCJZIl0sWzIsMCwiWCJdLFswLDEsIkMnIl0sWzIsMSwiXFxiZWdpbnthcnJheX17Y31DXFxcXFxcUGlfMV4xLVxcdGV4dHtjb21wbGV0ZX1cXGVuZHthcnJheX0iXSxbMCwxLCJmIl0sWzIsMCwiXFxzdWJzZXRlcSIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzMsMSwiXFxzdWJzZXRlcSIsMix7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV1d + \begin{tikzcd} + Y && X \\ + {C'} && {\begin{array}{c}C\\\Pi_1^1-\text{complete}\end{array}} + \arrow["f", from=1-1, to=1-3] + \arrow["\subseteq", hook, from=2-1, to=1-1] + \arrow["\subseteq"', hook, from=2-3, to=1-3] + \end{tikzcd} Let $X = 2^{\Q} \supseteq \WO$. We have already show that $\WO$ is $\Pi^1_1$-complete. + Set $\phi(x) \coloneqq \otp(x)$ ($\otp\colon \WO \to \Ord$ denotes the order type). We show that this is a $\Pi^1_1$-rank. + }{% + It suffices to show this for a $\Pi^1_1$-complete set. + We show that $\phi \coloneqq \otp$ is a + $\Pi^1_1$-rank for $\WO \subseteq 2^{\Q}$. + } Define $E \subseteq \Q^{\Q} \times 2^{\Q} \times 2^{\Q}$ by - \begin{IEEEeqnarray*}{rCl} - (f,x,y) \in E &:\iff& f \text{ order embeds $(x, \le_{\Q})$ to $(y,\le _{\Q})$}\\ - &\iff& \forall p,q \in \Q.~(p,q \in x \land p <_{\Q} q \implies f(p), f(q) \in y \land f(p) <_{\Q} f(q)) + \begin{IEEEeqnarray*}{Cl} + & (f,x,y) \in E\\ + :\iff& f \text{ order embeds $(x, \le_{\Q})$ to $(y,\le _{\Q})$}\\ + \iff& \forall p,q \in \Q.~(p,q \in x \land p <_{\Q} q \implies f(p), f(q) \in y \land f(p) <_{\Q} f(q)) \end{IEEEeqnarray*} - $E$ is Borel as a countable intersection of clopen sets. + $E$ is Borel\gist{ as a countable intersection of clopen sets}{}. Define - $x <^\ast_{\phi}$ + $x <^\ast_{\phi} y$ iff \begin{itemize} \item $(x, <_{\Q})$ is well ordered and @@ -51,8 +56,9 @@ This is equivalent to \begin{itemize} \item $x \in \WO$ and - \item $\forall f \in \Q^\Q.~(f,y,x) \not\in E$. + \item $\forall f \in \Q^\Q.~(f,y,x) \not\in E$, \end{itemize} + so it is $\Pi^1_1$. Furthermore $x \le_\phi^\ast y \iff$ either $x <^\ast_\phi y$ or @@ -61,12 +67,11 @@ i.e.~either $x<^\ast_\phi y$ or $x,y \in \WO$ and any order embedding of $(x,<_{\Q})$ to $(y, <_{\Q})$ is cofinal% - \footnote{% + \gist{\footnote{% Recall that $A \subseteq (x,<_{\Q})$ is \vocab{cofinal} if $\forall t \in x.~\exists a \in A.~t\le _{\Q} a$.% - } - in $(y, <_\Q)$ is - cofinal in $(y, <_{\Q})$ and vice versa. + }}{} + in $(y, <_\Q)$ and vice versa. Equivalently, either $(x <^\ast_\phi y)$ or \begin{IEEEeqnarray*}{rCl} @@ -76,7 +81,6 @@ \end{IEEEeqnarray*} \end{proof} - \begin{theorem} \label{thm:uniformization} Let $X$ be Polish and $R \subseteq X \times \N$ by $\Pi^1_1$ @@ -88,7 +92,6 @@ \] We say that $R^\ast$ \vocab[uniformization]{uniformizes} $R$.% \footnote{Wikimedia has a \href{https://upload.wikimedia.org/wikipedia/commons/4/4c/Uniformization_ill.png}{nice picture.}} - \end{theorem} \begin{proof} Let $\phi\colon R \to \Ord$ @@ -112,27 +115,28 @@ and \[ \bigcup_{n \in \N} C_n^\ast = \bigcup_{n \in \N} C_n. - \] + \] \end{corollary} \begin{proof} Define $R \subseteq X \times \N$ by setting - $(x,n) \in R :\iff x \in C_n$ + $(x,n) \in R :\iff x \in C_n$ and apply \yaref{thm:uniformization}. \end{proof} Let $X$ be a Polish space. -If $(X, \prec)$ is well founded (i.e.~there are no infinite descending chains) -then we define a rank $\rho_{y}\colon X > \Ord$ -as follows: +If $(X, \prec)$ is well-founded% +\gist{ (i.e.~there are no infinite descending chains)}{} +then we define a rank $\rho_{y}\colon X \to \Ord$ as follows: For minimal elements the rank is $0$. -Otherwise set $\rho_<(x) \coloneqq \sup \{\rho_<(y) + 1 : y \prec x\}$. +Otherwise set $\rho_\prec(x) \coloneqq \sup \{\rho_\prec(y) + 1 : y \prec x\}$. Let $\rho(\prec) \coloneqq \sup \{\rho_{\prec}(x) + 1 : x \in X\}$. -\begin{exercise} - $\rho(\prec) \le |X|^+$ (successor cardinal). - (for countable $<$) - \todo{TODO} -\end{exercise} +% \begin{exercise} +% $\rho(\prec) \le |X|^+$ (successor cardinal). +% (for countable $<$) +% \todo{TODO} +% % TODO QUESTION +% \end{exercise} \begin{theorem}[Kunen-Martin] \yalabel{Kunen-Martin}{Kunen-Martin}{thm:kunenmartin} If $(X, \prec)$ is wellfounded @@ -140,80 +144,85 @@ Let $\rho(\prec) \coloneqq \sup \{\rho_{\prec}(x) + 1 : x \in X\}$. then $\rho(\prec) < \omega_1$. \end{theorem} \begin{proof} + % TODO GIST + % TODO QUESTION where did we use analytic? Wlog.~$X = \cN$. There is a tree $S$ on $\N \times \N \times \N$ - (i.e.~$S \subseteq \cN^3$) - such that + (i.e.~$S \subseteq (\N \times \N \times \N)^{<\N}$) + such that \[ - \forall x, y \in \cN.~\left(x \succ y \iff \exists \alpha \in \N.~(x,\alpha,y) \in [S]\right). - \] + \forall x, y \in \cN.~\left(x \succ y \iff \exists \alpha \in \cN.~(x,\alpha,y) \in [S]\right). + \] Let \[ W \coloneqq \{w = (s_0,u_1,s_1,\ldots, u_n, s_n) : s_i, u_i \in \N^{n} \land (s_{i-1}, u_i, s_i) \in S\}. - \] + \] - So $|W| \le \aleph_0$. + Clearly $|W| \le \aleph_0$. Define $\prec^\ast$ on $W$ by setting \[(s_0,u_1,s_1,\ldots, u_n,s_n) \succ (s_0',u_1', s_1', \ldots, u_m', s_m') :\iff\] \begin{itemize} - \item $n < m$, - \item $\forall i \le n.~s_i \subsetneq s_i'$ and - \item $\forall i \le n.~u_i \subsetneq u_i'$. + \item $n < m$ and + \item $\forall i \le n.~s_i \subsetneq s_i' \land u_i \subsetneq u_i'$. + %\todo{$\subseteq$ or $\subsetneq$?} % TODO QUESTION \end{itemize} \begin{claim} $\prec^\ast$ is well-founded. \end{claim} \begin{subproof} - If $w_n = (s_0^n, u_1^n, \ldots, u_n^n, s_n^n)$ - was descending, - then let - \[ - x_i \coloneqq \bigcup s_i^n \in \cN - \] - and - \[ - \alpha_i \coloneqq \bigcup_n u_i^n \cN. - \] - We get $(x_{i-1}, \alpha_i, x_i) \in [S]$, - hence $x_{i-1} \succ x_i$ for all $i$, - but this is an infinite descending chain - in the original relation $\lightning$ + \gist{% + If $w_n = (s_0^n, u_1^n, \ldots, u_n^n, s_n^n)$ + was descending, + then let + \[ + x_i \coloneqq \bigcup s_i^n \in \cN + \] + and + \[ + \alpha_i \coloneqq \bigcup_n u_i^n \cN. + \] + We get $(x_{i-1}, \alpha_i, x_i) \in [S]$, + hence $x_{i-1} \succ x_i$ for all $i$, + but this is an infinite descending chain + in the original relation $\lightning$ + }{% + Use that $ \prec$ is well-founded. + } \end{subproof} + Hence $\rho(\prec^\ast) < |W|^+ \le \omega_1$. + % TODO QUESTION + We can turn $(X, \prec)$ into a tree $(T_\prec, \subsetneq)$ + with + \begin{IEEEeqnarray*}{rCl} + \rho(\prec) &=& \rho(T_{\prec}) + \end{IEEEeqnarray*} + by setting $\emptyset \in T_{\prec}$ + and + $(x_0,\ldots,x_n) \in T_\prec$,$x_i \in X =\cN$, + iff $x_0 \succ x_1 \succ x_2 \succ \ldots \succ x_n$. - \todo{Fix typos and end proof} -% Hence $\rho(\prec^\ast) < \omega_1$. -% -% We can turn $(X, \prec)$ into a tree $(T_\prec, \subsetneq)$ -% with -% \begin{IEEEeqnarray*}{rCl} -% \rho(\prec) &=& \rho(T_{\prec}) -% \end{IEEEeqnarray*} -% by setting $\emptyset \in T_{\prec}$ -% and -% $(x_0,\ldots,x_n) \in T_\prec$,$x_i \in X =\cN$, -% iff $x_0 \succ x_1 \succ x_2 \succ \ldots \succ x_n$. -% -% For all $x \succ y$ -% pick $\alpha_{x,y} \in \cN$ -% such that $(x, \alpha_{x,y}, y) \in [S]$ -% define -% \begin{IEEEeqnarray*}{rCl} -% \phi\colon T_{\prec} &\longrightarrow & W \\ -% \phi(x_0,x_1,\ldots,x_n) &\longmapsto & (x_0\defon{n}, \alpha_{x_0}, x_1\defon{n},\ldots, -% \alpha_{x_{n-1}}, x_n\defon{n}). -% \end{IEEEeqnarray*} -% Then $\phi$ is a homomorphism of $\subsetneq$ to $<^\ast$ -% and -% \[ -% \rho(<) = \rho(T_{\prec}, \subsetneq) \le \rho(<^\ast) < \omega_1. -% \] - -\todo{Exercise} + For all $x \succ y$ + pick $\alpha_{x,y} \in \cN$ + such that $(x, \alpha_{x,y}, y) \in [S]$ + define + \begin{IEEEeqnarray*}{rCl} + \phi\colon T_{\prec} \setminus \{\emptyset\} &\longrightarrow & W \\ + \phi(x_0,x_1,\ldots,x_n) &\longmapsto & (x_0\defon{n}, \alpha_{x_0}, x_1\defon{n},\ldots, + \alpha_{x_{n-1}}, x_n\defon{n}). + \end{IEEEeqnarray*} + Then $\phi$ is a homomorphism of $\subsetneq$ to $\prec^\ast$ + so + \[ + \rho(\prec) + = \rho(T_{\prec} \setminus \{\emptyset\} , \subsetneq) + \le \rho(\prec^\ast) + < \omega_1. + \] \end{proof} diff --git a/inputs/lecture_15.tex b/inputs/lecture_15.tex index ba7c57c..6f3f804 100644 --- a/inputs/lecture_15.tex +++ b/inputs/lecture_15.tex @@ -1,5 +1,8 @@ \lecture{15}{2023-12-05}{} +% TODO ANKI-MARKER + + \begin{theorem}[Boundedness Theorem] \yalabel{Boundedness Theorem}{Boundedness}{thm:boundedness}